Consider the following function on $[0,1]$: $$f(x)=\begin{cases}1&x\in\{a_1,a_2,\dots,a_n\}\\0 &\text{otherwise}\end{cases}$$ The $a_i$ are fixed and all in $[0,1]$. Determine whether $f$ is Riemintegrable or not.
By the Riemann–Lebesgue criterion $f$ is Riemann integrable. How to prove directly?
Take a partition $P$ of $[0,1]$ that is defined as: $$ P = \left\{0,a_1-\frac{\epsilon}{2},a_1+\frac{\epsilon}{2},\dots,a_n-\frac{\epsilon}{2},a_n+\frac{\epsilon}{2},1\right\}. $$ Clearly, $U(P,f) = \epsilon N$ and $L(P,f) = 0$. Since for any given $\epsilon>0$, a partition that could yield $$ U(P,f) - L(P,f) < \epsilon $$ can be found, $f$ is Riemann integrable.
To compute $\int f$, note that $\int f = \underline{\int f} = \sup_{P}L(P,f) = 0$, since for any given partition $P$, $L(P,f) = 0$.
HINT
You can bound the Riemann sum $I(P,\{t^*_i\})$ by $$0 \le I(P,\{t^*_i\}) \le n|P| $$ where $|P|$ is the length of the longest interval in the partition.
